An alpha particle of mass 4.00 u, moving to the right at 1.00x10^6 m/s, collides

Question

An alpha particle of mass 4.00 u, moving to the right at 1.00x10^6 m/s, collides with a proton of mass 1.00 u that is at rest before the collision. Find the speed of each particle after the collision, assuming a perfectly elastic collision.
(1u=1 atomic mass unit=1.67x10^-27 kg)
m1=6.68x10-27kg m2= 1.67x10-27kg
I set this up the way the book set up elastic collisions,
(6.68x10^-27)(1x10^6)=(1.67x10^-27)(0)
however, this ends up equalling 0?
the answer in the book is 6.0x10^5 m/s, 1.6x10^6 m/s
I'm not sure I started this right, thanks!

Explanation / Answer

initial momentum of total system = 4*10^6 + 0 let the velocity of alpha particle after collision be u and velocity of proton be v conservation of momentum: total initial momentum = total final momentum 4 x 10^6 = 4u + v for elastic collision coefficient of restitution =1 =( v-u)/(10^6-0) v-u = 10^6 v = u+ 10^6 plug this in first equation 4 x 10^6 = 5u + 10^6 u = .6 x 10^6 m/s v = 1.6 x 10^6 m/s

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